1. ## Chain Rule?

Given that $x = e^t$, find $\frac{dy}{dx}$ in terms of $\frac{dy}{dt}$ and show that

$\frac{d^2y}{dx^2} = e^{-2}\left(\frac{d^2y}{dt^2} -\frac{dy}{dt}\right)$.

EDIT: Sorry, It is a typo, it should be $\frac{d^2y}{dx^2} = e^{-2t}\left(\frac{d^2y}{dt^2} -\frac{dy}{dt}\right)$.

2. I remember doing this question!
*dives into the past papers*

I also remember doing it wrong and teacher correcting me

I can do the first part though...

Using the chain rule, we know:
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{dy}{dt} . \frac{dt}{dx}$

$x=e^t \\ \Longrightarrow \\ \frac{dx}{dt}=e^t \\ \Longrightarrow \\ \frac{dt}{dx}=\frac{1}{e^t}$

and if you substitute that in, you'll get this:

$\frac{dy}{dx}=\frac{1}{e^t}.\frac{dy}{dt}$

3. Originally Posted by Silver

I remember doing this question!
*dives into the past papers*

I also remember doing it wrong and teacher correcting me

I can do the first part though...

Using the chain rule, we know:
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{dy}{dt} . \frac{dt}{dx}$

$x=e^t \\ \Longrightarrow \\ \frac{dx}{dt}=e^t \\ \Longrightarrow \\ \frac{dt}{dx}=\frac{1}{e^t}$

and if you substitute that in, you'll get this:

$\frac{dy}{dx}=\frac{1}{e^t}.\frac{dy}{dt}$
Yes, I got that too. But it's the next stage that is tricky. How do I differentiate that:

$\frac{dy}{dx}=\frac{1}{e^t}.\frac{dy}{dt}$

?

4. Originally Posted by Air
Yes, I got that too. But it's the next stage that is tricky. How do I differentiate that:

$\frac{dy}{dx}=\frac{1}{e^t}.\frac{dy}{dt}$

?
$\frac{d^2 y}{dx^2} = \frac{d}{dx} \left( e^{-t} \, \frac{dy}{dt} \right)$

$= e^{-t} \frac{d}{dx} \left( \frac{dy}{dt} \right) + \frac{dy}{dt} \frac{d}{dx} \left( e^{-t} \right)$

using the product rule

$= e^{-t} \frac{d}{dt} \left( \frac{dy}{dt} \right) \, \frac{dt}{dx}+ \frac{dy}{dt} \, \frac{d}{dt} \left( e^{-t} \right) \frac{dt}{dx}$

using the chain rule

$= e^{-t} \, \frac{d^2y}{dt^2} \, e^{-t} + \frac{dy}{dt} \, \left(-e^{-t} \right) \, e^{-t}$

$= e^{-2t} \, \frac{d^2y}{dt^2} - \frac{dy}{dt} \, e^{-2t} .....$

5. Hi,

I'm not really sure because I can't get to your equation...
And sorry for the "d", it takes too long time to straighten them...

${\color{red}\frac{dy}{dx}}=e^{-t} \frac{dy}{dt}$

You're looking for $\frac{d^2y}{dx^2}$

-----------------------

\begin{aligned} \frac{d^2y}{dx^2} &=\frac{d}{dx} \left({\color{red}\frac{dy}{dx}}\right) \\ \\
&=\frac{d}{\color{green}dx} \left(e^{-t} \frac{dy}{dt}\right) \end{aligned}

Now, $x=e^t \implies {\color{green}dx}=e^t dt$ (we have to do it because the variable is t in the brackets, not x).

Replacing :

\begin{aligned} \frac{d^2y}{dx^2} &=\frac{d}{\color{green}dx} \left(e^{-t} \frac{dy}{dt}\right) \\
&=\frac{d}{e^t dt} \left(e^{-t} \frac{dy}{dt}\right) \\
&=e^{-t} \frac{d}{dt} \left(e^{-t} \frac{dy}{dt} \right)
\end{aligned}

Using the product rule :

\begin{aligned} \frac{d^2y}{dx^2} &=e^{-t} \left(-e^{-t} \frac{dy}{dt}+e^{-t} \frac{d^2y}{dt^2} \right) \\
&=e^{-2t} \left(-\frac{dy}{dt}+\frac{d^2y}{dt^2}\right) \end{aligned}

Edit : oh I see, you made a typo, Air

Re-edit : too fast for me Mr F

6. I think You have written the second part wrong. It should be:
$
\frac{d^2y}{dx^2} = e^{-2t}\left(\frac{d^2y}{dt^2} -\frac{dy}{dt}\right)
$

Solution:
$
\frac{dy}{dx}=\frac{1}{x}\frac{dy}{dt}
$

$
\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\ right)=\frac{d}{dx}\left(\frac{1}{x}\frac{dy}{dt}\ right)=\frac{1}{x}\frac{d}{dx}\frac{dy}{dt}-\frac{1}{x^2}\frac{dy}{dt}=\frac{1}{x}\frac{d}{dt} \frac{dy}{dx}-\frac{1}{x^2}\frac{dy}{dt}=\frac{1}{x^2}\frac{d^2y }{dt^2}-\frac{1}{x^2}\frac{dy}{dt}
$

$
=e^{-2t}\left(\frac{d^2y}{dt^2} -\frac{dy}{dt}\right)
$

Regards,